THIS MONTH'S PUZZLER
We use a centrifugal pump for tank truck loading. It's driven by a standard fixed-speed induction motor operating at 3,600 rpm. We have no controls on pump flow. The pump runs out on the pump curve during tank loading. An upset condition could cause a very high pressure at the pump suction. The piping system can handle the pressure but we have a concern about relief valve capacity downstream of the pump. The relief valve and downstream piping were designed based on flow being limited by pump capacity. But suction pressure can rise so high that the flow rate, if the pump weren't present, would exceed pump run-out capacity. Can the pump motor be forced to speed up to accommodate a higher flow rate by high pump suction pressure? Or will the motor and pump act as a brake on the system?
CONSIDER TWO IMPROVEMENTS
If, during an upset, the suction pressure increases so high that the pump is not necessary, the fixed-speed induction motor will run essentially at the same speed (3,600 rpm). The pump impeller will add kinetic energy and the volute will convert the kinetic energy to head pressure. The discharge pressure will be the sum of the suction pressure and the head developed by the pump. In hydraulic sense the pump cavity, the block valve and the piping represent the net resistance to the flow. The excessive discharge pressure can cause severe splashing at the tank and may pose a safety risk.
Consider the following improvements to reduce this risk:
1. Install a relief valve on the suction side. There is one present at the discharge side. However, if the upset is instantaneous, then the relief valve may not be much help — you need to make sure piping (suction and discharge) is adequate for the expected pressure.
2. If feasible consider throttling the discharge valve all the time; of course, this will slow down loading of the tank truck. However, it will help you keep on the pump curve operating range (the pump won't run off the curve). During an upset, however, you will have higher pressure than what you get with the valve wide open.
G.C. Shah, HSE project manager
Mustang Engineers, Sunnyvale, Calif.
FREE-WHEELING WILL OCCUR
As the suction pressure rises it will force flow through the pump and the motor current will fall since the suction pressure is doing the work originally performed by the motor. If the suction pressure continues to rise the flow will increase and start free-wheeling the pump shaft above its nominal speed. At that point the pump motor will become a generator and will start producing current in an attempt to slow down the pump. So the short answer to your question is, "Yes, the motor will act as a brake until it finally blows the overloads in the starter and trips off line. At that point the pump will likely free-wheel and do little to restrict the flow into the downstream piping."
Hunter Vegas, senior project manager
Avid Solutions, Inc., Winston Salem, N.C.
THE MOTOR BECOMES A GENERATOR
A three-phase induction motor will only spin at (or very near) the rated rpm when energized. If shaft work required actually does go below zero, the motor will act as a brake and produce electrical power rather than consume it.
Chris Rentsch, process engineer
Dow Chemical, Midland, Mich.
PUMP DAMAGE MAY OCCUR
As the pump inlet pressure increases due to upset, the liquid flows to discharge side at increasing rate. (It was assumed pump discharge side is at lower pressure compared to suction side due to safety valve opening.) This is the case of a hydraulic turbine, where the liquid head drop across the impeller converts into mechanical work. This is in addition to the electrical energy supplied to the pump motor as there were no controls specified. Therefore in such a case, the pump shaft can acquire very high speeds depending the differential pressure. Depending on the properties of fluid handled, high pressure fluid depressurization across the pump may cause severe cavitation and damage to the pump.
THIS MONTH'S PUZZLER